Question

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Answer 1

I hope this helps you

Answer 2

18.) Given,
A=30°
To prove,
(i)2 cos²A-1=1-2 sin²A=cos 2A
(ii)cos 2A=cos ²A-sin²A=(1-tan²A)÷ (1+tan²A)
ATQ,
(i) 2 * (cos30°)²-1=2*(√3/2)²-1
=2*3/4-1
=3/2-1
=1/2
Now,
1-2(sin30°)²=1-2*(1/2)²
=1-2*1/4
=1-1/2
=1/2
Again,
cos(2*30°)=cos60°
=1/2
So,
2(cos 30°)²-1=1-2(sin 30°)²=cos(2×30°)

(ii)cos(2*30°)=cos 60°
=1/2
Now,
(cos 30°)²-(sin 30°)²=(√3/2)²-(1/2)²
=3/4-1/4
=2/4=1/2
Again,
1-(tan30°)²/1+(tan30°)²=1-(1/√3)²/1+(1/√3)²
=(1-1/3)/(1+1/3)
=(2/3)/(4/3)
=(2*3)/(4*3)
=2/4=1/2
So,
cos(2*30°)=(cos30°)²-(sin30°)²=1-(tan30°)²÷1+(tan30°)²

19.) Given,
A=30°
To verify,
tan2A=(2tanA)÷(1-tan²A)
ATQ
tan2A=tan(2*30°)=tan(60°)=√3
Now,
(2*tan30°)÷{1-(tan30°)²}=
(2*1/√3)÷{1-(1/√3)²}=
2/√3÷{1-1/3}=
(2√3/3)÷(2/3)=
(2√3*3)/(3*2)=
6√3/6=√3
So,
tan2A=(2tanA)÷(1-tan²A)


Hope this helps you!